LintCode-Maximum Subarray Difference
Given an array with integers.
Find two non-overlapping subarrays A and B, which |SUM(A) - SUM(B)| is the largest.
Return the largest difference.
Note
The subarray should contain at least one number
Example
For [1, 2, -3, 1], return 6
Challenge
O(n) time and O(n) space.
Solution:
1 public class Solution { 2 /** 3 * @param nums: A list of integers 4 * @return: An integer indicate the value of maximum difference between two 5 * Subarrays 6 */ 7 public int maxDiffSubArrays(ArrayList<Integer> nums) { 8 int len = nums.size(); 9 if (len==0) return 0; 10 11 int[] leftMin = new int[len]; 12 int[] leftMax = new int[len]; 13 int endMin = nums.get(0), endMax = nums.get(0); 14 leftMin[0] = endMin; 15 leftMax[0] = endMax; 16 for (int i=1;i<len;i++){ 17 //Calculate max subarray. 18 endMax = Math.max(nums.get(i), nums.get(i)+endMax); 19 leftMax[i] = Math.max(leftMax[i-1],endMax); 20 21 //Calculate min subarray. 22 endMin = Math.min(nums.get(i),nums.get(i)+endMin); 23 leftMin[i] = Math.min(leftMin[i-1],endMin); 24 } 25 26 int[] rightMin = new int[len]; 27 int[] rightMax = new int[len]; 28 endMin = nums.get(len-1); 29 endMax = nums.get(len-1); 30 rightMin[len-1] = endMin; 31 rightMax[len-1] = endMax; 32 for (int i=len-2;i>=0;i--){ 33 endMax = Math.max(nums.get(i),nums.get(i)+endMax); 34 rightMax[i] = Math.max(rightMax[i+1],endMax); 35 endMin = Math.min(nums.get(i),nums.get(i)+endMin); 36 rightMin[i] = Math.min(rightMin[i+1],endMin); 37 } 38 39 int maxDiff= 0; 40 for (int i=0;i<len-1;i++){ 41 if (maxDiff<Math.abs(leftMin[i]-rightMax[i+1])) 42 maxDiff = Math.abs(leftMin[i]-rightMax[i+1]); 43 44 if (maxDiff<Math.abs(leftMax[i]-rightMin[i+1])) 45 maxDiff = Math.abs(leftMax[i]-rightMin[i+1]); 46 } 47 return maxDiff; 48 49 50 } 51 }
